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I've tried a few "criteria" to check if this is irreducible. According to Maple it only has one entirely real root which I suspect is not rational but I can't prove it so I'm attempting to check if $p$ is irreducible.

Eisenstein's Criterion doesn't work here and I'm yet to find a suitable transformation such that it could work. I also read that if a polynomial is irreducible over $\Bbb F_q$, with $q$ a prime not dividing the leading coefficient, then it is irreducible over $\Bbb Q$ so I reduced the polynomial modulo $2$ to obtain

$$p \equiv x^5 + x^3 + 1 \mod 2.$$

I think this is correct but then I need to know how to check the irreducibility of this new polynomial over $\Bbb F_2$. Do I simply need to check that neither $0$ nor $1$ are roots of this polynomial? (And am I applying this theorem correctly?)

If this polynomial IS irreducible over $\Bbb Q$, is the splitting field obtained by simply adjoining the roots to $\Bbb Q$?

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  • $\begingroup$ Since the polynomial is of degree $5$ it might be possible for it to have no roots in $\mathbb{Q}$, but be reducible. Writing it as a product of polynomials of degree $2$ and $3$ might be possible. $\endgroup$
    – Stefan4024
    Apr 9 '17 at 11:11
  • $\begingroup$ To answer your last question: yes. This follows straight from the definition of a splitting field. $\endgroup$
    – user583416
    Apr 9 '17 at 11:18
  • $\begingroup$ Oh, I've found something wonderful. Answer coming up. $\endgroup$ Apr 9 '17 at 11:19
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.A little bit of scouting for nice irreducibility criteria throws up some very nice results:

Here is a lovely lemma by (Prof.) Ram Murty:

Let $f(x) = a_mx^m + ... + a_1x + a_0$ be a polynomial of degree $m$ in $\mathbb Z[x]$. Let $H = \displaystyle\max_{0 \leq i \leq m-1} \left|\frac{a_i}{a_m}\right|$. If $f(n)$ is prime for some $n \geq H+2$, then $f(x)$ is irreducible in $\mathbb Z[x]$.

I'll give the link : http://cms.dm.uba.ar/academico/materias/2docuat2011/teoria_de_numeros/Irreducible.pdf

In our case, $a_m = 1$, and the maximum of all the quantities in question is $2$. Hence, if $f(n)$ is prime for some $n \geq 4$, then we are done.

You can check that for $n=4$, the number $f(4) =919$, which is prime!

Hence, it follows that the polynomial is irreducible.


ASIDE : There is also a "shifted" base (base shifts from $0...n-1$ to $|b| < \frac n2$) version of Cohn's criteria, which will tell you that if $f(10)$ is prime, then the given polynomial is irreducible. This matches that description, since all coefficients are between $-5$ and $5$. Very interestingly, $f(10) = 98779$ is also prime! (Hence, another proof by another wonderful result).

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  • $\begingroup$ That is very cool! Thanks for the link. $\endgroup$ Apr 9 '17 at 11:28
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    $\begingroup$ You are welcome. I've seen some of these criteria very long ago, while doing a Galois theory polynomial "scouting mission". $\endgroup$ Apr 9 '17 at 11:31
  • $\begingroup$ Nice method. Just wondering, why not going already with $f(4)=919$ which is also a prime? $\endgroup$
    – Sil
    Apr 26 '18 at 23:30
  • $\begingroup$ You are right, it works out. I was doing that calculation mentally, so I skipped directly to $n=5$ because I must have made some mistake at $f(4)$. $\endgroup$ Apr 26 '18 at 23:34
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By Gauss' Lemma we have that the polynomial is irreducible over $\mathbb{Q}$ if and only if it's irreducible over $\mathbb{Z}$. Now the easiest way would be to prove that polynomial is irreducible over $\mathbb{Z}_2$, which would be enough.

Assume it's reducible. As the polynomial has no roots over $\mathbb{Z}_2$, then the only possibility is if it's a product of polynomials of degree $2$ and $3$. So assume that:

$$x^5 + x^3 + 1 = (x^3 + ax^2 + bx + c)(x^2 + dx + e) \quad \quad \text{over } \quad\mathbb{Z}_2$$

Then multiply everything out and compare the factors. As $a,b,c,d,e \in \mathbb{Z}_2$, you only have two options. Eventually you will get that $c=e=1$ and $a=d=b$. But this would imply that $c+bd + ea = a^2 + a + 1 = 1$ in $\mathbb{Z}_2$. But this is impossible, as it's the coefficient in front of $x^2$ and it should be $0$.

Hence the polynomial is irreducible over $\mathbb{Z}_2$ and eventually $\mathbb{Z}$ and $\mathbb{Q}$

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    $\begingroup$ I think I'll accept this one as being the most useful in a general setting, thanks a lot. $\endgroup$ Apr 9 '17 at 11:28
  • $\begingroup$ One could also just divide by the only irreducible polynomial in $\mathbb{Z_2}[x]$ which is $x^2+x+1$ and see it cannot be the factor. $\endgroup$
    – Sil
    Aug 11 '18 at 9:23
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    $\begingroup$ @Sil True. However I doubt that an OP asking factorizing of a polynomial is aware that $x^2+x+1$ is the only irreducible quadratic in $\mathbb{Z}_2$ $\endgroup$
    – Stefan4024
    Aug 11 '18 at 9:25
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Your proof is almost complete: it is indeed sufficient to prove that $p(x)\in \mathbb F_2[x]$ (the reduction mod $2$ of your polynomial) is irreducible.
The polynomial $p(x)$ has no factor of degree $1$ since it has no zero in $\mathbb F_2$, so there remains only to prove that $p(x)$ has no factor which is an irreducible polynomial $g(x)\in \mathbb F_2[x]$ of degree $2$.
But the only such irreducible polynomial is $g(x)=x^2+x+1$ and long division proves that it does not divide $p(x)$. All is proved.

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By the rational root theorem, any rational root of $p$ would have to be a divisor of the constant term, so $\pm1$. Clearly, these are not roots.

Similary, any quadratic factor would have to be $x^2+ax\pm1$ with $a\in\Bbb Z$. You might be able to exclude these manually ...

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